Harristotelian Logic

As a hobby I've been playing around with FLT partly because it is considered a hopeless quest. That way, I knew I could probably play with it forever.

However, I just stumbled upon something that seems profound to me so I thought I'd toss it here so that someone give it a quick critique. This is not long.

Given all that usual stuff about x^n + y^n = z^n

let x=af, y=bg, z=ch where x+y=h^n or n^{n-1}h^n, z-x=g^n or n^{n-1}g^n,

and z-y=f^n or n^{n-1}f^n. And, (x+y-z)^n = n(z-x)(z-y)(x+y)Q where Q is all those other terms that are hard to write for the general case. For n=3 it is one. Also I have Q=q^n

Trying z divisible by n and using the above I get the following which must be true.

f^n + 2nfghq + g^n = n^{n-1}h^n subtracting f^n + nfgp + g^n = (f+g)^n

gives

nfg(p-2hq) = (f+g)^n - n^{n-1}h^n

p is always divisible by (f+g) because nfgp=(f+g)^n - (f^n + g^n)

since (f+g) must be divisible by n, because z is divisible by n, requires that h or q be divisible by n which is a contradiction.

The same comes up with x or y divisible by n since you get

g^n + 2nfghq - h^n = n^{n-1}f^n subtracting

g^n + nghp - h^n = (g-h)^n

gives

ngh(p-2fq) = (g-h)^n - n^{n-1}f^n

which requires that f or q be divisible by n because (g-h) is divisible by n which is a contradiction.

For n>3 with neither x,y or z divisible by n the above is just as conclusive. Therefore, I'm stuck wondering what could be wrong with the above only one of you will tell me.

Supposing that z is divisible by n, then z=nch and x+y=n^{n-1}h^n